Answer
$\approx 1.083$
Work Step by Step
The formula for the arc length is
$s=\int_{a}^b \sqrt{1 + (y')^2} dy$
Differentiate $y=\frac{x^3}{6}+ \frac{1}{2x}$ with respect to x.
$y'=\frac{x^2}{2}- \frac{1}{2x^2}$
Substitute the value of y' in $1 + (y')^2$
$1+(y')^2=1+[\frac{1}{2}(x^2 - x^{−2})]^2$
$=\frac{1}{4}(x^2 + x^{−2})^2$
Substitute the value of $1 + (y')^2$ in the formula for the arc length $s=\int_{a}^b \sqrt{1 + (y')^2} dy$
Solve for the distance s
$s=\int_{0}^2\sqrt{\frac{1}{4}(x^2 + x^{-2})}^{2} dx$
$=\int_{0}^2 \frac{1}{2}(x^2 - x^{-2})dx$
$=\frac{1}{2}[x^2 - x^{-2}]_0^2\ $
$= \frac{13}{12}
$\approx 1.083$
